Question
imageshack(.)(us)/photo/my-images/834/j6wv.jpg/
imageshack(.)(us)/photo/my-images/17/6n4x.jpg/
please PLEASE IM BEGGING YOU MY TEACHER WILL BE GIVING ME 79 IF I DIDN'T GET THIS .. OR ELSE LOWER THAN THAT. PLEASE
imageshack(.)(us)/photo/my-images/17/6n4x.jpg/
please PLEASE IM BEGGING YOU MY TEACHER WILL BE GIVING ME 79 IF I DIDN'T GET THIS .. OR ELSE LOWER THAN THAT. PLEASE
Answers
BARBIE LEE
just delete the parethesis please
Reiny
The key to this question is to realize that all the right -angled triangles are similar, and have the
ratio of 3:4:5 in the length of their sides.
Mark the two angles in triangle BCD with o and x
( you should be able to see that since XD | BC , so you can mark XDC as o and DCX as x
continue through the other triangles...)
Triangle XDC is similar to triangle DCB
XD:XC:DC = DC:DB:CB = 4:3:5
4p:3p:4
4p/4 = 4/5
4p = 16/5 ----> XD = 16/5
3p/4 = 3/5
3p=12/5 ----> XC = 12/5
and of course ---> CD = 4
now you have all the sides of triangle XDC
Did you notice that
3:4:5 = (5/4)(12/5 : 16/5 : 4) ??
continue in this fashion until you have all sides needed.
ratio of 3:4:5 in the length of their sides.
Mark the two angles in triangle BCD with o and x
( you should be able to see that since XD | BC , so you can mark XDC as o and DCX as x
continue through the other triangles...)
Triangle XDC is similar to triangle DCB
XD:XC:DC = DC:DB:CB = 4:3:5
4p:3p:4
4p/4 = 4/5
4p = 16/5 ----> XD = 16/5
3p/4 = 3/5
3p=12/5 ----> XC = 12/5
and of course ---> CD = 4
now you have all the sides of triangle XDC
Did you notice that
3:4:5 = (5/4)(12/5 : 16/5 : 4) ??
continue in this fashion until you have all sides needed.
Anonymous
yes sir ive got that already. i don't know what to do next. please help